Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
int g[maxn][maxn];
int main(int argc, char const *argv[]) {
  int n, m;
  scanf("%d %d", &n, &m);
  int mn = 505;
  int nice = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      scanf("%d", &g[i][j]);
      if (g[i][j] != 0) nice = 0;
      if (i == 1) {
        mn = min(mn, g[i][j]);
      }
    }
  }
  if (nice == 1) {
    printf("0\n");
    return 0;
  }
  vector<pair<int, int>> v;
  for (int i = 0; i <= mn; ++i) {
    vector<vector<int>> a(n + 1, vector<int>(m + 1));
    for (int j = 1; j <= m; ++j) {
      a[1][j] = i;
    }
    vector<pair<int, int>> ops;
    for (int j = 0; j < i; ++j) ops.push_back({1, 1});
    for (int j = 1; j <= m; ++j) {
      if (g[1][j] > a[1][j]) {
        int sum = max(0, g[1][j] - a[1][j]);
        for (int k = 1; k <= n; ++k) {
          a[k][j] += sum;
        }
        for (int k = 0; k < sum; ++k) ops.push_back({2, j});
      }
    }
    for (int j = 2; j <= n; ++j) {
      int sum = max(0, g[j][1] - a[j][1]);
      for (int k = 1; k <= m; ++k) {
        a[j][k] += sum;
      }
      for (int k = 0; k < sum; ++k) ops.push_back({1, j});
    }
    bool good = 1;
    for (int j = 1; j <= n; ++j) {
      for (int k = 1; k <= m; ++k) {
        if (a[j][k] != g[j][k]) {
          good = 0;
          break;
        }
      }
      if (!good) break;
    }
    if (good) {
      if (v.empty() or v.size() > ops.size()) {
        v = ops;
      }
    }
  }
  if (v.empty())
    printf("-1\n");
  else {
    printf("%d\n", (int)v.size());
    for (auto i : v) {
      if (i.first == 1)
        printf("row %d\n", i.second);
      else
        printf("col %d\n", i.second);
    }
  }
  return 0;
}